Question

The graph between $1/\lambda$ and stopping potential $(V)$ of three metals having work functions ${\varphi }_1,{\varphi }_2$ and ${\varphi }_3$ in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following option(s) is/are correct?
(Here, $\lambda$ is the wavelength of the incident ray)

• A. Ratio of work function ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$
• B. Ratio of work function ${\varphi }_1:{\varphi }_2:{\varphi }_3=4:2:1$
• C. $\tan \theta$ is directly proportional to $hc/e$, where $h$ is Planck's constant and $c$ is the speed of light
• D. The violet colour light can eject photoelectrons from metals 2 and 3 .

Answer 1

Answer: (A), (C)

$\frac{hc}{\lambda }-\varphi =eV$
$V=\frac{hc}{e\lambda }-\frac{\varphi }{e}$
For plate I
$\frac{{\varphi }_1}{hc}=0.001$
Plate 2
$\frac{{\varphi }_2}{hc}=0.002$
Plate 3
$\frac{{\varphi }_3}{hc}=0.004$
${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$
For plate $2$ , threshold wavelength
$\lambda =\frac{hc}{{\varphi }_2}=\frac{hc}{0.002hc}=\frac{1000}{2}=500nm$
For plate $3$ , threshold wavelength
$\gamma =\frac{hc}{{\varphi }_3}=\frac{hc}{0.004hc}=\frac{1000}{4}=250nm$
Since, violet colour light will eject photoelectrons from plate 2$notfrom$3$ .